[FOM] NBG, Subsets and Cantor's Theorem
Thomas Forster
T.Forster at dpmms.cam.ac.uk
Wed Apr 4 11:25:10 EDT 2007
Dear Paul,
The answer to your question is: yes. You do indeed need some form of
the axiom of separation. This is shown by the fact that there are
consistent set theories in which Cantor's theorem canot be proved:
Church's 1976 system for one. Such systems have weakened versions
of the separation scheme.
On Tue, 3 Apr 2007, Studtmann, Paul wrote:
> I have a question about the way in which one proves Cantor¹s Theorem that
> there is no one to one function from a set onto its power-set within NBG. I
> should say that I am taking as my example of NBG the presentation that
> Mendelson gives in his Introduction to Mathematical Logic, third edition.
> My question is simply: is the axiom of subsets needed in order to prove
> Cantor¹s Theorem within NBG?
>
> I will state briefly why I came to the thought that it might be, though I
> would not be at all surprised to learn that my mathematical reasoning has
> gone awry somewhere it normally does. I shall also try to keep the
> reasoning as informal as possible.
>
> Any variable or constant that is in lower case refers to a set.
>
> The axiom of subsets states that the intersection, I, of any class, X, and
> any set, y, is a set.
>
> Suppose there is a 1-1 function, f, from the set, a, onto the power-set of
> a, P(a). Then, there exists a class, C, such that for all sets, y, y E C if
> and only if y E a and it is not the case that y E f(y), where 'E' is the
> membership relation. (By the comprehension schema)
> Now, on the assumption that C is in P(a), a contradiction eventually follows
> in the familiar way. But how does one prove that C is in P(a)? Well, as far
> as I can tell, what one needs to do is prove that C is a subset of a. This
> requires proving that (i) every member of C is in a, which follows easily
> from the defining condition of C; and (ii) C is a set. Why does one have to
> prove that C is a set? Because if C is not a set, then it could be the case
> that everything that is in it is in the set, a, even though C is not in the
> power-set of a the power-set axiom in NBG says that every sub-set, x, of
> another set, y, is in the power-set of y, not that every sub-class, X, of
> another set, y, is in the power-set of y. Now, with the axiom of subsets,
> the fact that C is a set follows easily from the facts that (i) the
> intersection of C and a is C, and (ii) a is a set. But without the axiom of
> subsets I do not see how one would derive the fact that C is a set. And so
> that has led me to my present question.
>
>
> I would not be surprised to find out that I am making some kind of mistake
> here my grasp on mathematical fact is flimsy at best. But I would very
> much like to know the answer to the question. So if anyone could either
> tell me the answer or cite some relevant source, I would be very much
> obliged.
>
>
> Paul Studtmann
>
>
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